Problem: You have found the following ages (in years) of all 5 porcupines at your local zoo: $ 8,\enspace 2,\enspace 25,\enspace 14,\enspace 6$ What is the average age of the porcupines at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 porcupines at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{8 + 2 + 25 + 14 + 6}{{5}} = {11\text{ years old}} $ Find the squared deviations from the mean for each porcupine. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $8$ years $-3$ years $9$ years $^2$ $2$ years $-9$ years $81$ years $^2$ $25$ years $14$ years $196$ years $^2$ $14$ years $3$ years $9$ years $^2$ $6$ years $-5$ years $25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{9} + {81} + {196} + {9} + {25}} {{5}} $ $ {\sigma^2} = \dfrac{{320}}{{5}} = {64\text{ years}^2} $ The average porcupine at the zoo is 11 years old. The population variance is 64 years $^2$.